Hello to everyone,
Could somebody please tell me how do we solve exercises like the one I describe below:
$\displaystyle exp(1+(O(1/n))^2)$ = e + O(1/n)
Thanks a lot,
Tasos
First of all, I want to thank you mmmbot for answering.
Sorry if I didn't write it well. By writing exp I meant e to the power.
So I mean: [e to the power of (1 + (O(1/n))^2)] = e + O(1/n)
P.S.1: Could you tell me how to write it in the post so as it will appear correctly?
P.S.2: What you say is right: exp(1) and e are the same thing. That is what I mean.
I'm checking to see whether or not O(1/n) is function notation ... be right back.
A big "Oh, please excuse me".
I thought O and n were constants. (I just saw the big Oh function for the first time at Wikipedia.)
No, you are correct. O(1/n) is good.
I cannot help you with this exercise.
Now, I read it as follows.
e^[1 + O(1/n)^2] = e + O(1/n)
Again, sorry for goofing up your thread.
~ Mark
Are you sure you put the parentheses at the right place? Because this is correct, but not optimal, as you shall see:
I'll apply usual properties of the "big O" notation. If you don't know them, just ask, the proofs are very short.
First you have $\displaystyle \left(O\left(\frac{1}{n}\right)\right)^2=O\left(\f rac{1}{n^2}\right)$, and it is more usual to write it this way.
Then $\displaystyle e^{1+O\left(\frac{1}{n^2}\right)}=e e^{O\left(\frac{1}{n^2}\right)}= e\left( 1+O\left(\frac{1}{n^2}\right)\right)$ $\displaystyle =e + e\, O\left(\frac{1}{n^2}\right)=e+O\left(\frac{1}{n^2} \right)$.
I used the following expansion of the exponential at 0: $\displaystyle e^{O(u)}=1+O(u)$ when $\displaystyle u$ tends to 0, composed with the sequence $\displaystyle \left(\frac{1}{n^2}\right)_n$ which tends to 0.
Now, $\displaystyle 0\leq\frac{1}{n^2}\leq \frac{1}{n}$, hence the previous big O can be replaced by $\displaystyle O\left(\frac{1}{n}\right)$, and this is what you need.
Let us now suppose that the question was $\displaystyle e^{\left(1+O\left(\frac{1}{n}\right)\right)^2}=1+O \left(\frac{1}{n}\right)$, which seems more plausible to me.
Then we would have: $\displaystyle \left(1+O\left(\frac{1}{n}\right)\right)^2=1+O\lef t(\frac{1}{n}\right)+O\left(\frac{1}{n^2}\right)=1 +O\left(\frac{1}{n}\right)$ (as before, the second big O can be included in the first one), and by the same computation as above we conclude $\displaystyle e^{\left(1+O\left(\frac{1}{n}\right)\right)^2}=e+O \left(\frac{1}{n}\right)$.
ps: about the way to write symbols, there's a section in the forum related to "LaTeX", and this is where you should look for help about this.
This is because of this result:
In fact, as soon as a function $\displaystyle f$ is differentiable at $\displaystyle 0$, we have $\displaystyle f(x)=f(0)+O(x)$ as $\displaystyle x$ tends to 0 (because $\displaystyle \frac{f(x)-f(0)}{x}\to f'(0)$ so that $\displaystyle \frac{f(x)-f(0)}{x}=O(1)$ as $\displaystyle x\to0$).
And if $\displaystyle g(u)=O(u)$ as $\displaystyle u\to0$, then $\displaystyle g(u)\to_{u\to 0} 0$, so that we can compose: $\displaystyle f(g(u))=f(0)+O(g(u))=f(0)+O(u)$. This can be written $\displaystyle f(O(u))=f(0)+O(u)$ as $\displaystyle u$ tends to 0.
Now, in this expansion, you can replace $\displaystyle u$ by any sequence which converges to 0. For instance, $\displaystyle f(O\left(\frac{1}{n^2}\right))=f(0)+O\left(\frac{1 }{n^2}\right)$. If $\displaystyle f=\exp$, you get what I wrote and used.
(I'm thinking of something that may have been confusing: when I wrote $\displaystyle e\left( 1+O\left(\frac{1}{n^2}\right)\right)$, it meant $\displaystyle e\times \left( 1+O\left(\frac{1}{n^2}\right)\right)$, not exponential of the parenthesis)